描述
Given a string s and a character letter, return the percentage of characters in s that equal letter rounded down to the nearest whole percent.
Example 1:
Input: s = "foobar", letter = "o"
Output: 33
Explanation:
The percentage of characters in s that equal the letter 'o' is 2 / 6 * 100% = 33% when rounded down, so we return 33.
Example 2:
Input: s = "jjjj", letter = "k"
Output: 0
Explanation:
The percentage of characters in s that equal the letter 'k' is 0%, so we return 0.
Note:
1 <= s.length <= 100
s consists of lowercase English letters.
letter is a lowercase English letter.
解析
根据题意，给定一个字符串 s 和一个字符 letter ，返回 s 中包含 letter 的数量占 s 长度的整数百分比。
其实这道题就是一个考察简单的字符串、小数计算以及小数取整的题目，没有什么好说的，直接使用内置函数 count 找 s 中 letter 的数量即可，然后计算所占长度的百分比取整即可。
时间复杂度为 O(N) ，空间复杂度为 O(N) 。
解答
class Solution(object):
def percentageLetter(self, s, letter):
"""
:type s: str
:type letter: str
:rtype: int
"""
return int(1.0*s.count(letter)/len(s)*100)
运行结果
85 / 85 test cases passed.
Status: Accepted
Runtime: 40 ms
Memory Usage: 13.3 MB
原题链接
您的支持是我最大的动力
